Problem 31 - College Physics

Question

A 0.300 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.80 \(\mathrm{m} / \mathrm{s}\) when it makes a head-on collision with a stationary 0.150 \(\mathrm{kg}\) glider. (a) Find the magnitude and direction of the final velocity of each glider if the collision is elastic. (b) Find the final kinetic energy of each glider.

Step-by-Step Solution

Answer

(a) Glider 1: 0.32 m/s right, Glider 2: 1.12 m/s right, (b) KE Glider 1: 0.0154 J, KE Glider 2: 0.0944 J.

Step 1: Define the Initial Conditions

The initial velocity of the 0.300 kg glider is \( v_{1i} = 0.80 \, \text{m/s} \) (moving to the right), and the 0.150 kg glider is initially stationary, so \( v_{2i} = 0 \, \text{m/s} \).

Step 2: Apply Conservation of Momentum

For an elastic collision, momentum is conserved. Thus, we have the equation \( m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \), where \( m_1 = 0.300 \, \text{kg} \), \( m_2 = 0.150 \, \text{kg} \), \( v_{1f} \) and \( v_{2f} \) are final velocities of the gliders.

Step 3: Apply Conservation of Kinetic Energy

Since the collision is elastic, kinetic energy is also conserved. The equation is \( \frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2 = \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2 \).

Step 4: Solve the Equations

Use the equations from Step 2 and Step 3 to solve for \( v_{1f} \) and \( v_{2f} \). You will find that \( v_{1f} = 0.32 \, \text{m/s} \) and \( v_{2f} = 1.12 \, \text{m/s} \).

Step 5: Determine Direction of Velocities

The positive direction is to the right. Since both final velocities are positive, both gliders continue moving to the right.

Step 6: Calculate Final Kinetic Energy of Each Glider

The final kinetic energy of the 0.300 kg glider is \( KE_{1f} = \frac{1}{2} \times 0.300 \, \text{kg} \times (0.32 \, \text{m/s})^2 = 0.0154 \, \text{J} \). The final kinetic energy of the 0.150 kg glider is \( KE_{2f} = \frac{1}{2} \times 0.150 \, \text{kg} \times (1.12 \, \text{m/s})^2 = 0.0944 \, \text{J} \).

Key Concepts

Conservation of Momentum

In physics, the law of conservation of momentum plays a crucial role in understanding and analyzing collisions. Momentum, defined as the product of an object's mass and its velocity, is a conserved quantity in a closed system, meaning the total momentum of the system before a collision is equal to the total momentum after the collision.
If we consider two gliders on a frictionless track, one of mass 0.300 kg and the other 0.150 kg, the total momentum prior to the collision can be expressed using the formula:

• Total initial momentum: \( m_1 v_{1i} + m_2 v_{2i} \)

Where:

• \( m_1 = 0.300 \, \text{kg} \), \( v_{1i} = 0.80 \, \text{m/s} \)
• \( m_2 = 0.150 \, \text{kg} \), \( v_{2i} = 0 \, \text{m/s} \)

Since the 0.150 kg glider is stationary initially, its contribution to the initial total momentum is zero. Post-collision, this same total momentum manifests differently distributed among the final velocities \( v_{1f} \) and \( v_{2f} \). Solving this ensures the principle of momentum conservation is satisfied, critical for determining these velocities.

Conservation of Kinetic Energy

In an elastic collision, not only is momentum conserved, but kinetic energy is also conserved. Kinetic energy is the energy an object possesses due to its motion, given by the formula \( KE = \frac{1}{2} m v^2 \).

For the collision between the two gliders, the total kinetic energy before the collision is entirely due to the moving glider. This can be formulated as:

• Total initial kinetic energy: \( \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 \)

Substituting the initial conditions, we get:

• \( \frac{1}{2} \times 0.300 \, \text{kg} \times (0.80 \, \text{m/s})^2 + \frac{1}{2} \times 0.150 \, \text{kg} \times (0 \, \text{m/s})^2 \)

After the collision, this total kinetic energy is spread over the two gliders, emphasizing the requirement for conserving energy alongside momentum. Solving these concurrent equations ensures the correct distribution of velocities and energies, reflecting the physical laws governing elastic collisions.

Final Velocities Calculation

To determine the final velocities \( v_{1f} \) and \( v_{2f} \) of the gliders after an elastic collision, we employ the conservation principles of both momentum and kinetic energy as derived:

• Momentum conservation equation: \( m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \)
• Kinetic energy conservation equation: \( \frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2 = \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2 \)

First, substitute the known values into both equations. The complexity arises due to the simultaneous nature of these equations, but solving them systematically yields:

• \( v_{1f} = 0.32 \, \text{m/s} \)
• \( v_{2f} = 1.12 \, \text{m/s} \)

This indicates that post-collision, the heavier glider slows down, while the lighter, previously stationary glider moves faster. A simple check for compatibility with the conservation principles can confirm these values are accurate.

Kinetic Energy Calculation

Calculating the kinetic energy of each glider after the collision provides insight into the dynamics of the interaction. The formula used to find the kinetic energy is: \( KE = \frac{1}{2} m v^2 \).

For the first glider (0.300 kg moving at \( 0.32 \, \text{m/s} \)), its final kinetic energy \( KE_{1f} \) can be computed as:

• \( KE_{1f} = \frac{1}{2} \times 0.300 \, \text{kg} \times (0.32 \, \text{m/s})^2 \approx 0.0154 \, \text{J} \)

Similarly, calculate the final kinetic energy for the second glider (0.150 kg, moving at \( 1.12 \, \text{m/s} \)):

• \( KE_{2f} = \frac{1}{2} \times 0.150 \, \text{kg} \times (1.12 \, \text{m/s})^2 \approx 0.0944 \, \text{J} \)

Combining these energies gives the total kinetic energy post-collision, aligning it with the total initial energy to validate the elastic nature of the event. Each glider's movement reflects the redistribution of energy, maintaining overall system equilibrium.